More generally, it can be shown rigorously that \(q\le \kappa _\star \)  [1]}. The corresponding dominant eigenvector, denoted by \(s_j\) , also admits a simple perturbative expression [2]}:
\(s_j = \delta _{j\star } + \frac{\kappa _{j\star }}{\kappa _j - \kappa _\star } + \sum _{\ell \ne \star } \frac{\kappa _{j\ell }\kappa _{\ell \star }}{(\kappa _j - \kappa _\star )(\kappa _\ell - \kappa _\star )}+\ldots \) 
